Question

Circular cooling fins of diameter \(D=1 \mathrm{~mm}\) and length $L=30 \mathrm{~mm}\(, made of copper \)(k=380 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$, are used to enhance heat transfer from a surface that is maintained at temperature \(T_{s 1}=132^{\circ} \mathrm{C}\). Each rod has one end attached to this surface \((x=0)\), while the opposite end \((x=L)\) is joined to a second surface, which is maintained at \(T_{s 2}=0^{\circ} \mathrm{C}\). The air flowing between the surfaces and the rods is also at \(T_{\infty}=0^{\circ} \mathrm{C}\). and the convection coefficient is $h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. (a) Express the function \(\theta(x)=T(x)-T_{\infty}\) along a fin, and calculate the temperature at \(x=L / 2\). (b) Determine the rate of heat transferred from the hot surface through each fin and the fin effectiveness. Is the use of fins justified? Why? (c) What is the total rate of heat transfer from a \(10-\mathrm{cm}\) by \(10-\mathrm{cm}\) section of the wall, which has 625 uniformly distributed fins? Assume the same convection coefficient for the fin and for the unfinned wall surface.

Short answer

Question: Determine the temperature at the midpoint (\(x=L/2\)) along the fin. Answer: To find the temperature at \(x=L/2\), substitute this value into the derived temperature distribution equation obtained in step 3 of the solution. Make sure to convert the value of \(x\) to meters before calculation.

Step-by-step solution

Define the given parameters

Here, we have been given the following parameters for the circular cooling fins: - Diameter \(D = 1 \mathrm{~mm}\) - Length \(L = 30 \mathrm{~mm}\) - Thermal conductivity of copper \(k = 380 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Surface temperatures: \(T_{s 1}=132^{\circ} \mathrm{C}\) and \(T_{s 2}=0^{\circ} \mathrm{C}\) - Free air flow temperature \(T_{\infty}=0^{\circ} \mathrm{C}\) - Convection coefficient \(h = 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Calculate the temperature distribution along the fin

In order to find the temperature distribution along the fin, we will use the fin equation: \(\frac{d^{2}\theta(x)}{dx^{2}}+\frac{hP}{k A_c}\theta(x)=0\) Where \(\theta(x)=T(x)-T_{\infty}\), \(A_c\) is the cross-sectional area of the fin, and \(P\) is the perimeter of the fin. The fin equation can be re-written as: \(\frac{d^{2}\theta(x)}{dx^{2}}+m^2\theta(x)=0\) Where \(m =\sqrt{\frac{hP}{k A_c}}\) We solve this for \(\theta(x)\), giving us a general equation for the temperature distribution along the fin: \(\theta(x) = C_1 e^{-mx} + C_2 e^{mx}\) Now we need to apply the boundary conditions to solve for \(C_1\) and \(C_2\): At \(x = 0\), \(T(x) = T_{s1}\), so \(\theta(0) = 132^{\circ}\mathrm{C}\) At \(x = L\), \(T(x) = T_{s2}\), so \(\theta(30\mathrm{~mm}) = 0^{\circ}\mathrm{C}\)

Find the constants \(C_1\) and \(C_2\)

By applying the boundary co... conditions, we can now solve for the constants \(C_1\) and \(C_2\): At \(x=0\), \(\theta(0)=132^{\circ} \mathrm{C}\) gives us: \(132 = C_1 e^{-m \cdot 0} + C_2 e^{m \cdot 0}\) which simplifies to: \(132 = C_1 + C_2\) At \(x=L\), \(\theta(30\mathrm{~mm})=0^{\circ} \mathrm{C}\) gives us: \(0 = C_1e^{-30m}+ C_2e^{30m}\) Now solving this system of equations (for \(C_1\) and \(C_2\)), we obtain the temperature distribution equation for the given fin.

Calculate the temperature at \(x=L/2\)

After obtaining the constants \(C_1\) and \(C_2\) from the previous step, we now have the equation for the temperature distribution along the fin. To find the temperature at \(x=L/2\), substitute this value into the derived temperature distribution equation. Remember to convert the value of \(x\) to meters before calculation.

Determine the rate of heat transfer through each fin and fin effectiveness

The rate of heat transfer through each fin can be calculated using the following equation: \(q_{fin} = kA_c\frac{d\theta(x)}{dx}\)(evaluated at \(x=0\)) Fin effectiveness (\(\eta_{fin}\)) can be determined using the following relation: \(\eta_{fin} = \frac{q_{fin}}{q_{max}}\) where \(q_{max}=hA_s\theta(0)\) and \(A_s\) is the surface area of the fin. To find \(q_{fin}\) and \(\eta_{fin}\), calculate the derivatives of the temperature distribution equation and evaluate them at \(x=0\). Then, use the effectiveness formula to determine if the use of fins is justified.

Calculate the total heat transfer rate

To determine the total heat transfer rate from a \(10 \mathrm{~cm} \times 10 \mathrm{~cm}\) section of the wall with 625 uniformly distributed fins, we first need to find the heat transfer rate through the unfinned surface and then add it to the heat transfer rate through the 625 fins. The total heat transfer rate \(q_{total}\) can be calculated as: \(q_{total} = (A_{wall} - 625A_s)h\theta(0) + 625q_{fin}\) Calculate the value of \(q_{total}\) using the given values and found values for \(q_{fin}\) and \(\eta_{fin}\).